Two Inlet Pipes and a Leaking Tank
A municipal water tank is filled by two inlet pipes. The larger pipe alone takes 5 minutes less than the smaller pipe to fill the empty tank. When both inlet pipes are opened together, the empty tank is filled in 6 minutes. A leak at the bottom of the tank can empty a full tank in 30 minutes. The caretaker wants to know whether the two inlet pipes can fill the tank while the leak remains open, and how the filling time changes if the smaller pipe is closed after 4 minutes. The tank capacity is not given, so the solution must work with rates. Assume constant flow rates, that the leak rate is independent of depth, and that the tank starts empty. The caretaker also rules out the smaller pipe taking less than 5 minutes alone, since that would make the larger pipe's time negative.
1. Find the time each pipe alone would take to fill the empty tank.
Read the worked answer
Let the smaller pipe take x minutes alone; then the larger takes x − 5 minutes. In one minute they fill 1/x and 1/(x−5) of the tank. Together:
1/x + 1/(x−5) = 1/6.
Multiplying by 6x(x−5): 6(x−5) + 6x = x(x−5), so 12x − 30 = x² − 5x, hence x² − 17x + 30 = 0.
Factorising: (x − 15)(x − 2) = 0. x = 2 is rejected because x − 5 would be negative, which is impossible for a filling time. Thus x = 15 minutes for the smaller pipe and 10 minutes for the larger pipe.
2. With both inlet pipes and the leak open, find the time to fill the tank and decide whether it fills at all.
Read the worked answer
Rates: smaller 1/15, larger 1/10, leak −1/30 per minute. Net rate = 1/15 + 1/10 − 1/30 = 2/30 + 3/30 − 1/30 = 4/30 = 2/15 per minute.
Time to fill = 1 ÷ (2/15) = 15/2 = 7.5 minutes. Since the net rate is positive, the tank does fill despite the leak.
3. If the smaller pipe is closed after 4 minutes, find the total time needed to fill the tank.
Read the worked answer
With both inlets and the leak open for 4 minutes, fraction filled = 4 × (2/15) = 8/15. Remaining fraction = 7/15.
After the smaller pipe is closed, the rate is larger pipe minus leak = 1/10 − 1/30 = 3/30 − 1/30 = 2/30 = 1/15 per minute.
Time for remaining part = (7/15) ÷ (1/15) = 7 minutes. Total time = 4 + 7 = 11 minutes.